Question

Area and the perimeter of right angled triangle are 6cm square and 12cm respectively. Find the sides of the right angled triangle.

Answer 1

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a+b+5=12

a+b+5=12a+b=7⇒a=7−b...(1)

a+b+5=12a+b=7⇒a=7−b...(1)a

a+b+5=12a+b=7⇒a=7−b...(1)a 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b)

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =25

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =25

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0(b−4)(b−3)=0

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0(b−4)(b−3)=0⇒b=4,3

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0(b−4)(b−3)=0⇒b=4,3a=7−b⇒a=3,4

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0(b−4)(b−3)=0⇒b=4,3a=7−b⇒a=3,4Area=

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0(b−4)(b−3)=0⇒b=4,3a=7−b⇒a=3,4Area= 2

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0(b−4)(b−3)=0⇒b=4,3a=7−b⇒a=3,4Area= 21

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0(b−4)(b−3)=0⇒b=4,3a=7−b⇒a=3,4Area= 21

a+b+5=12a+b=7⇒a=7−b...(1)a 2 +b 2 =5 2 a 2 +b 2 =25...(2)From (1) and (2)(7−b) 2 +b 2 =2549+b 2 −14b+b 2 =252b 2 −14b+24=0b 2 −7b+12=0(b−4)(b−3)=0⇒b=4,3a=7−b⇒a=3,4Area= 21 ×4×3=6sq. units